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During a trip on an expressway, Don drove a total of x

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During a trip on an expressway, Don drove a total of x [#permalink] New post 17 Nov 2006, 15:37
During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

- 8.5%
- 50%
- (x/12)%
- (60/x)%
- (500/x)%
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 [#permalink] New post 17 Nov 2006, 16:23
(T1 - T2)/T2 * 100 = [{5/30 + (x-5)/60} - x/60] / x/60 * 100
= 500/x%
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[#permalink] New post 17 Nov 2006, 17:15
During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?

- 8.5%
- 50%
- (x/12)%
- (60/x)%
- (500/x)%

Time (5/30) + (x-5)/60= (10+x-5)/60=(5+x)/60

Time with avg 60 = x/60

(5+x)/60 - x/60 = 5/60

(5/60)*100/ (x/60) = 500%/x
Answer E
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 [#permalink] New post 19 Nov 2006, 18:40
Remaining journey: (x-5)

So:
First 5 miles -> 1/6 hour
remaining -> (x-5)/60 hour
Total = [10 + (x-5)]/60 = x+5/60

If entire journey on constant 60 mph, then time = x/60

Extra time needed = 5/60 hour -> 1/12 hour

Percentage = 1/12 * 60/x * 100% = 500/x %
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 [#permalink] New post 20 Nov 2006, 00:12
Could someone say why picking numbers does not work?

suppose x=10 miles

Then....

30/60*t=5
1/2t=5
t=6 minutes for the first 5 miles

and

60/60*t=5
t=5 minutes for the remaining 5 miles

Total 5+6=11 minutes

if he had travelled 60 per hour all the way;
1t=10
t=10 minutes......

So his travel time was 10% greater...

Include 10 instead of x in answer choices and look for 10% return.... what is wrong here?
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 [#permalink] New post 30 Dec 2006, 19:32
I see that it is too late. I happen to come across only today.

SimaQ wrote:
Could someone say why picking numbers does not work?

suppose x=10 miles

Then....

30/60*t=5
1/2t=5
t=6 minutes for the first 5 miles t should be 10

and

60/60*t=5
t=5 minutes for the remaining 5 miles

Total 5+6=11 minutes Then total = 5+10=15

if he had travelled 60 per hour all the way;
1t=10
t=10 minutes......

So his travel time was 10% greater... so 50% greater which is 500/10%, so picking nos. too work here
Include 10 instead of x in answer choices and look for 10% return.... what is wrong here?
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Re: Problem Solving [#permalink] New post 30 Dec 2006, 20:05
i'm not getting what others are.
i solved this way.

it is only for the 5 mile section that speeds are different,namely 30 and 60 miles per hour.
for the remainder,the speed is same in both the cases.so the time spent will be same for remainder of the distance.

for the 5 mile section,t1 = 5/30 hrs
t2 = 5/60 hrs

hence %age increase = (t1 - t2)/t1 * 100=50 %

what is wrong ????
and what is the oa??
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Re: Problem Solving [#permalink] New post 30 Dec 2006, 23:45
AK47 wrote:
i'm not getting what others are.
i solved this way.

it is only for the 5 mile section that speeds are different,namely 30 and 60 miles per hour.
for the remainder,the speed is same in both the cases.so the time spent will be same for remainder of the distance.

for the 5 mile section,t1 = 5/30 hrs
t2 = 5/60 hrs

hence %age increase = (t1 - t2)/t1 * 100=50 %

what is wrong ????
and what is the oa??


t2 should be (x-5)/60 hs.

% increase = ((5/30 + (x-5)/60) - x/60) / (x/60) *100%

= 5/x *100% = 500/x%.
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Re: Problem Solving [#permalink] New post 31 Dec 2006, 01:00
i misread the problem.answer is choice E.I agree.
Re: Problem Solving   [#permalink] 31 Dec 2006, 01:00
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